1326. Minimum Number of Taps to Open to Water a Garden - Detailed Explanation

Problem Statement

You are given a garden represented as a one-dimensional interval ([0, n]) and an integer array ranges of length ( n + 1 ). Each element ranges[i] indicates that if you open the tap at position ( i ), it can water the interval ([i - \text{ranges}[i],\ i + \text{ranges}[i]]) (with the interval boundaries clipped to ([0, n])). Your task is to determine the minimum number of taps you need to open so that the entire garden is watered. If it is impossible to water the entire garden, return (-1).

For example, if:

• ( n = 5 )
• ( \text{ranges} = [3,4,1,1,0,0] )

The tap at position 0 can water ([0 - 3, 0 + 3] \Rightarrow [0, 3]), the tap at position 1 can water ([1 - 4, 1 + 4] \Rightarrow [0, 5]), etc. In this case, the entire garden ([0, 5]) can be watered by just opening tap 1, so the answer is 1.

Hints

1. Convert to Intervals:
   Each tap ( i ) with range ranges[i] can be represented as an interval: [ \text{interval}_i = \Big[\max(0, i - \text{ranges}[i]),\ \min(n, i + \text{ranges}[i])\Big] ] The problem then becomes an interval covering problem: find the minimum number of intervals (taps) required to cover the interval ([0, n]).

2. Greedy / Jump Game Approach:
   Instead of sorting and then iterating over intervals, you can precompute, for every starting position in the garden, the farthest position reachable by any tap that can start watering at that position. Then use a greedy approach similar to the "Jump Game II" problem:

   • Let current_end be the current rightmost point you can reach with the taps you have opened.

   • Let next_end be the farthest point you can reach by considering all taps starting within the current segment.

   • Increment the tap count and update current_end to next_end until the entire garden is covered.

3. Edge Cases:
   Consider cases where:

   • The garden is already fully watered by one tap.

   • No tap can start watering from the very beginning (position 0), making it impossible to cover the garden.

   • Some taps may have a range of 0 and thus do not help extend the coverage.

Detailed Approach

Step 1: Preprocess Taps into Maximum Reach Array

Create an array maxReach of size ( n + 1 ) to record, for each starting position, the farthest right endpoint that can be reached by any tap whose watering interval begins at that position. For each tap ( i ) (from 0 to ( n )):

• Calculate the left boundary:
  [ \text{left} = \max(0, i - \text{ranges}[i]) ]

• Calculate the right boundary:
  [ \text{right} = \min(n, i + \text{ranges}[i]) ]

• Update the array:
  [ \text{maxReach[left]} = \max(\text{maxReach[left]},\ \text{right}) ]

This tells you that from position left, you can cover up to right using tap ( i ).

Step 2: Greedy Coverage Using the "Jump Game" Technique

Initialize:

• taps ( = 0 ) (the number of taps opened so far),
• current_end ( = 0 ) (the current rightmost point that is watered),
• next_end ( = 0 ) (the farthest point reachable by considering the next tap(s)).

Iterate over positions ( i ) from 0 to ( n ):

• Update next_end to be the maximum of its current value and maxReach[i].
• When you reach position ( i ) equal to current_end, it means you must open a new tap to extend the watering range:
  • Increment taps.
  • Set current_end ( = ) next_end.
  • If current_end reaches or exceeds ( n ), the garden is fully watered.

If at any point ( i ) exceeds next_end (i.e., you cannot extend the coverage), return (-1) because it's impossible to water the entire garden.

Step 3: Return the Result

After processing, if the entire garden ([0, n]) is covered, return the total number of taps opened; otherwise, return (-1).

Complexity Analysis

• Time Complexity:
  Preprocessing takes ( O(n) ) and the greedy loop takes ( O(n) ). Overall, the time complexity is ( O(n) ).

• Space Complexity:
  The extra space used is ( O(n) ) for the maxReach array.

Python Code